I am learning QFT. Earlier we showed that a complex field can be decomposed like so: $$\begin{align*} \phi &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\big(a(\mathbf{k})e^{-ikx}+b^\dagger(\mathbf{k})e^{ikx}\big) \\ \phi^\dagger &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\big(b(\mathbf{k})e^{-ikx}+a^\dagger(\mathbf{k})e^{ikx}\big). \end{align*}$$ An exercise was given to me to show that the Feynman propagators $\langle 0 | T \phi(x)\phi(y) |0 \rangle$ and $\langle 0 | T \phi^\dagger(x)\phi^\dagger(y) |0 \rangle$ are both zero. Unfortunately, I'm not getting zero in my calculation, so what am I doing wrong?
This is what I have. First, $$ \begin{align*} \phi(y)|0\rangle &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}} b^\dagger(\mathbf{k})e^{iky} |0\rangle \\ &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} e^{iky} |\mathbf{k} \rangle . \\ \langle 0 | \phi(x) & = \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}} \langle 0| a(\mathbf{k})e^{-ikx} \\ &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} \langle \mathbf{k} | e^{-ikx} .\end{align*} $$ Then combining them, using $\langle \mathbf{k} | \mathbf{k'} \rangle = (2\pi)^3 2\omega_k \delta(\mathbf{k}-\mathbf{k}') $, and integrating over one of the $k$'s, $$ \langle 0 | T \phi(x)\phi(y) |0 \rangle = \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\omega_k}e^{ik(y-x)} $$ assuming that $x^0>y^0$. But if $x^0 < y^0$, then it seems we can just switch $y$ and $x$ (unless this is where the issue lies?).
I found a few sites saying something like that $\phi$ creates a particle but destroys an antiparticle, or was it the other way around? In any case, it is not clear to me how that is the case at all. Isn't $b^\dagger$ the creation operator for particles and $b$ the annihilation operator for antiparticles? I also fail to see how this answers why the propagator vanish.
